Triangle $\triangle ABC$ has a right angle at $C$, $\angle A = 60^\circ$, and $AC=10$.  Find the radius of the incircle of $\triangle ABC$.
Explanation: We begin by drawing a diagram:

[asy]
size(100);
pair A,B,C;
real x = sqrt(3);
C=(0,0); A=(10,0); B=(0,10*x);
draw(A--B--C--cycle);
draw(rightanglemark(B,C,A,30));
label("$A$",A,SE); label("$C$",C,SW); label("$B$",B,NW); label("10",(A+C)/2,S);

real r = 5*sqrt(3) - 5;
draw(Circle((r,r),r));
[/asy]

Since $\angle A = 60^\circ$, we have $\angle B = 180^\circ - 90^\circ - 60^\circ = 30^\circ$.  Then $\triangle ABC$ is a $30 - 60 - 90$ triangle, so $BC=AC\sqrt{3}=10\sqrt{3}$ and $AB=2AC=20$.  We can compute the area of $\triangle ABC$ as \[ [\triangle ABC] = \frac{1}{2}(AC)(BC)=\frac{1}{2}(10)(10\sqrt{3}) = 50\sqrt{3}.\]Let the incircle of $\triangle ABC$ have radius $r$.  A triangle with inradius $r$ and semiperimeter $s$ has \[\text{area} = rs,\]so we have  \[ [\triangle ABC] = r \left( \frac{10+10\sqrt{3}+20}{2} \right) = r(15+5\sqrt{3}).\]Setting these two area expressions equal gives \[50\sqrt{3}=r(15+5\sqrt{3}).\]Solving for $r$ gives  \[r = \frac{10\sqrt{3}}{3+\sqrt{3}} = \frac{10\sqrt{3}(3-\sqrt{3})}{9-3} = \boxed{5(\sqrt{3}-1)}.\]